I have two arrays:
arr1 = [1,2,3]
arr2 = [2,3]
What is the most handy way to check that all values from the arr2 contains in arr1.
If you just want to check if all of the elements from one list are present in the other list, you can simply use -- using the shorter list at the left side of the operator:
iex> [2, 3] -- [1, 2, 3]
[]
If all of the elements from the first list are present in the second one, the result should be an empty list.
Note, however, that this doesn't account for some cases, for instance, if your first list had a duplicate element, and the other list had the element, but just once, you wouldn't get an empty list:
iex> [2, 3, 3] -- [1, 2, 3]
[3]
But in that case, technically, the second list does not contain all of the elements from the first one.
If you would like to just check for presence of the elements, there are other simple solutions, like:
Enum.all?([2, 3, 3], &Enum.member?([1, 2, 3], &1))
Not sure about how efficient that is, though, as for each element of the first list you're checking if it's present in the second one (however, it will stop checking as soon as one element doesn't meet the condition, and the functions from Enum are typically optimized, so it might be good enough)
And yet another option, would be to use MapSet.
You can do:
MapSet.subset?(MapSet.new([2, 3]), MapSet.new([1, 2, 3]))
This will also work for duplicate elements, as MapSets work as sets, so you cannot have duplicate elements in them.
Related
Summary of problem
Ultimate goal
I would like to take a sub-array from a large input numpy array. This sub array is dynamic, and every iteration through the larger numpy input array will change the sub array so that I can perform a set of calculations that depend on previous iterations of the array. This involves nested for loops, which I realize is not very pythonic, but I don't know of another way.
Problem
The problem arises when I add to the existing dynamic sub-array, it seems to grow extra bracketing. This seems simple to fix, but I am having trouble adapting my Matlab knowledge of array indexing to numpy indexing. I have not even started implementing my calculations yet, but I cannot seem to get the structure of this loop correct.
What I've tried
I have [tried this originally in Pandas][1]. Originally, I thought I could write a pretty simply program to do this using pandas indexing and column naming. But it was SLOW! So I trying to streamline this by
changing the architecture and
relying on numpy instead of Pandas.
Below is a simple program that emulates what I want to do. I am sure I will have other questions, but this is the start. I have a simple (5, 2) array that I loop through the rows of. With each row after row 0, I add the new row to the top of the temp sub-array and delete the last row of the array, maintaining a (2, 2) array throughout. However, as you will see when you run this code, it results in some strange behavior that results in not being able to write the results into the output array. You will also see that I have tried several ways to add and delete columns. Whether these are optimal is besides the point - the current code is the closest I have gotten to running this program!
Some Example code
This code 'works' in the sense that it doesn't trow errors. However, it doesnt' produce the desired results. In this case it would be an output array with the same values as the inputs (because I am not doing any calculations- this is just to get the architecture correct). The desired result would be that each loop creates a sub array in this order:
n=1 [1 1]
n=2 [[1,1], [2,2]]
n=3 [[2, 2], [3, 3]]
n=4 [[3, 3], [4, 4]]
...
N [[N-1, N-1], [N, N]].
This does not need to be limited to 2 items (if list) or rows (if array), and the length will be set by an input variable. Thus, the size of this array must be dynamic (set during the call of the function). Furthermore, I supply a simple example here, but each loop will basically need to add a row from the input. It will be a little more advanced than simply a 2 member NDarray. Lists have the advantage of being able to use .append and .pop attributes, but as far as I can tell, arrays do not. I present the following code example using only arrays.
import numpy as np
a = np.array([[1, 1], [2, 2], [3, 3], [4,4], [5,5]])
print('Original a array: ', a)
out = np.empty_like(a)
b = np.empty(len(a[0,:]))
for ii, rr in enumerate(a):
if ii == 0:
c = [a[ii]]
else:
print('Before: ', c)
#Add next row from array a to the temp array for calculations
c = np.insert(c, 1, [rr], axis=0)
print('During: ', c)
#Remove the last row of the temp array prior to calculations
#indices_to_remove = [0]
#d = c[~np.isin(np.arange(c.size), [indices_to_remove])]
d = c[1::]
c = [d]
print('After: ', c)
#Add the temp array to the output array after calculations
#THIS THROWS ERRORS, AND I THINK IT IS DUE TO THE INCREASING NUMBERS OF BRACKETS.
#out[ii, :] = c
#print(c)
[1]: https://stackoverflow.com/questions/70186681/nested-loops-altering-rows-in-pandas-avoiding-a-value-is-trying-to-be-set-on?noredirect=1#comment124076103_70186681
MATLAB is 1-base indexing whereas Python uses 0-base indexing.
let's say we have a 2D array like this:
a= [[1, 2],
[3, 4],
[5, 6]]
In MATLAB if you do a(1, 1) you will get 1 in python a[1, 1] you will get 4. Also as you know in MATLAB you can do natural indexing which works if you do a(6) = 6 in python if you do that you will get IndexError: . it is almost the same, except in python it starts from 0.
Here is the working example with your desired results.
import numpy as np
a = np.array([[1, 1], [2, 2], [3, 3], [4,4], [5,5]])
test = []
for idx in range(len(a)):
if idx == 0:
test.append(a[idx])
test.append(a[idx:idx+2, :])
# remove the last [5, 5]
test.pop(-1)
for i in test:
print(i, end=',\n')
output
[1 1],
[[1 1]
[2 2]],
[[2 2]
[3 3]],
[[3 3]
[4 4]],
[[4 4]
[5 5]],
class Solution:
def transformArray(self, arr: List[int]) -> List[int]:
x=arr
while True:
f=True
for i in range(1,len(arr)-1):
if arr[i-1]<arr[i] and arr[i]>arr[i+1]:
f=False
x[i]=x[i]-1
print(x[i])
if arr[i-1]>arr[i] and arr[i]<arr[i+1]:
f=False
x[i]=x[i]+1
print(x[i])
#print(x)
x=arr
if f==True:
break
return x
In the above code both the if statements don't execute , only the second one does. I have tried using elif but it still doesn't work. What am i missing here?
For your code, I considered two types of examples as input for the array list
For example 1, when the arr = [1, 2, 1, 4], the 2nd element is bigger than 1st and 3rd
The first if statement (if arr[i-1]<arr[i] and arr[i]>arr[i+1]:) is working, because both the conditions are met and it gives the output x = [1, 1, 1, 4]
In example 2, when the arr = [3, 2, 3, 4], the 2nd element is smaller than the 1st and 3rd
The second if statement (if arr[i-1]>arr[i] and arr[i]<arr[i+1]:) is working, because both the conditions are met and it gives the output x = [3, 3, 3, 4]
So, the working of if statements largely depends on the elements in the array. Both the if statements' purpose is totally opposite. If one satisfies the condition the other will not.
Hope my answer provides some clarification.
let x = [1, 2, 2, 3, 3, 3, 1].reversed()
for element in x.method_name() {
print(element)
}
This returns
Value of type 'ReversedCollection<[Int]>' has no member 'method_name'.
Why? How do I reference the method I have created and have it do the functions I need it to do?
However, if I use the below, the problem seems to disappear. I would just like to pass in an array and do let the function do all, i.e.:
let x = Array([1, 2, 2, 3, 3, 3, 1].reversed())
Just in case you don't fully understand the motivation behind this overload of reversed returning a ReversedCollection instead of an Array, a ReversedCollection is just a "reversed view" of your original array. It is not a reversed copy of the original array. This is to save time and space, like a "lazy" collection. See this post for more details.
This is why you need the Array(...) initialiser to turn the reversed collection back into an array. You are opting out of the laziness.
On the other hand, there is another overload of reversed that returns an Array directly. Normally this overload is not selected because it is defined in a less specific type - Sequence, as opposed to Array. You need to give enough information about the type to use this overload:
let x: [Int] = [1, 2, 2, 3, 3, 3, 1].reversed()
I have a nested array:
arr = [[1,nil,2,3,4], [2,nil,4,5,6], [6,nil,3,3,5]]
Any elements at the same index in the subarrays that are nil across the array must be removed. The second index in all subarrays have nil.
I did this:
collection = arr.transpose.select(&:any?).transpose
# => [[1, 2, 3, 4], [2, 4, 5, 6], [6, 3, 3, 5]]
It works for me, albiet I am using transpose twice. Could this technique lead to data getting mixed up? It looks fool proof to me.
With the nil-vs-false caveat that #CarySwoveland noted in a comment, yes, your double-transpose is safe: it will only work on data that's rectangular to begin with, and it will produce equally-rectangular data as output. You're filtering out whole rows, so nothing can get misaligned.
While it's not super efficient, it's not too bad, and is far more expressive & readable than more direct looping & manipulation.
This is the array:
array = [ 1, 2, 3, [4, 5, 6] ]
Can I use "delete_at" method to delete the "5"?
array.delete_at[x] method
What would the correct syntax be?
Your 'array' has only 4 elements. If it's subarray you probably should do something like that
array[3].delete_at(1)
to delete the second element of subarray that's a fourth element of 'array' array.
Welcome to Stack Overflow!
This one is longer and less efficient but it allows you to select the item to be deleted by value instead of by position (array index). That's useful when you don't know the position.
array.map {|x| x.delete(5) if x.instance_of?(Array); x}